Question

$A$ sphere contracts in volume by $0.01$%, when taken to the bottom of sea $1km$ deep. The bulk modulus of the material of the sphere is (Given density of sea water may be taken as $1.0\times 10^{3}kg{m}^{-3})$.

• A. $4.9\times 10^{10}N$ $m^{-2}$
• B. $9.8\times 10^{10}N$ $m^{-2}$
• C. $4.9\times 10^{9}N$ $m^{-2}$
• D. $9.8\times 10^{9}N$ $m^{-2}$

Answer 1

Answer: (B)

Here, $\frac{\Delta V}{V}$ $=\frac{0.01}{100}$, $h=1km$ $=10^{3}m$
$\rho$ $=1.0\times 10^{3}kg$ $m^{-3}$
$p=h\rho g$ $=10^3\times 1\times 10^3\times 9.8$
$=9.8\times 10^{6}N$ $m^{-2}$
$\therefore$ $B=\frac{p}{\Delta V/V}$ $=\frac{9.8\times 10^6\times 100}{0.01}$ $=9.8\times 10^{10}N$ $m^{-2}$